momasri
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Posts posted by momasri
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booooooooooooo
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Can you try this PHP code:
<?php
$firstname = $_GET['firstname'];
$lastname = $_GET['lastname'];
echo 'Welcome to our website' . $$firstname . " " . $lastname;
?>
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The Ajax Code:
<script language="javascript" type="text/javascript"> <!-- //Browser Support Code function ajaxFunction(){ var ajaxRequest; // The variable that makes Ajax possible! try{ // Opera 8.0+, Firefox, Safari ajaxRequest = new XMLHttpRequest(); } catch (e){ // Internet Explorer Browsers try{ ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); } catch (e) { try{ ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); } catch (e){ // Something went wrong alert("Your browser broke!"); return false; } } } // Create a function that will receive data sent from the server ajaxRequest.onreadystatechange = function(){ if(ajaxRequest.readyState == 4){ var ajaxDisplay = document.getElementById('ajaxDiv'); ajaxDisplay.innerHTML = ajaxRequest.responseText; } } var pic = document.getElementById('pic').value; var queryString = "?pic=" + pic; ajaxRequest.open("GET", "ajax-example.php" + queryString, true); ajaxRequest.send(null); } //--> </script>The PHP code:
<?php require_once("configure.php"); mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable to connect to database" . mysql_error()); @mysql_select_db("$DBName") or die("Unable to select database $DBName" . mysql_error() ); $user_sqla = "select * from $emaila " ; while($user_array = mysql_fetch_array($user_resulta)) { $pictures = $user_array[picture]; ?> <form name='myForm'> <?php echo "<div align='left' style='padding-left:25px'><table border='1'><tr><td><img src=".$pictures." width='65' height='65'><input type='hidden' name='pic' id='pic' value=".$pictures.">". $pictures;?></td> <td><div style="padding-left:5px; padding-top:15px"><?php echo "<a href='#' onclick='ajaxFunction()' >Remove</a></td></tr></table></div>"; ?></div> </form><?php } ?>the ajax-example.php code:
<?php // Retrieve data from Query String $age = $_GET['pic']; echo $age; ?>
Now the Php code gives the following result:
"image1" "Remove"
"image2" "Remove"
"image3" "Remove"
when I click the "Remove" the result I keep getting is "image1" even if I click the remove beside image3 I still get the Image1 Name transffered from my ajax code to the ajax-example.php page
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sweet thanks, this solves the main problem. now I have a sub question from the above code:
in the Body of the page I write:
// this line displays the Pictures I have the "$dire" folder as specified in the code above
<?php getPictures(); ?>
// this line makes a Previous Page if exists
// and makes a Next button for Next Page views if they exist
<?php if ( $has_previous ) echo '<div align="left" style="padding-left:100"><a href="?page='.($page - 1).'">← Previous Page</a></div>'; if ( $has_next ) echo '<div align="right" style="padding-right:100"><a href="?page='.($page + 1).'"> Next Page →</a></p>'; ?>
// the code below gives me numbers for each page, so if the totall number of pictures per page is 5, and i have 20 pictures it gives me links 0 1 2 3 4
My question is How can I set my first page to be number "1" insted of number "0"
now I am sure the Fix for it is not in the code below, but I thought I would paste all so you can make use of this Picture Pigntaion
$directory = $dire .'/'; if (glob("$directory*.jpg") != false) { $filecount = count(glob("$directory*.jpg")); echo "You have " . $filecount . " Number of Files"; echo "<br/><br/>"; $total = $filecount / $per_page; echo "That means you have " .$total . " Number of Pages to view"; for ($i = 0; $i <= $total; $i++) if ($i >= 0) { echo '<p class="prev"> <a href="?page='.$i.'"> '.$i.' </a></p>'; } } ?> -
Actuly this is my first line few lines of code:
require_once("configure.php");
mysql_connect($DBhost,$DBuser,$DBpass) or die("Unable to connect to database" . mysql_error());
@mysql_select_db("$DBName") or die("Unable to select database $DBName" . mysql_error() );
$user_sql = "select institute from $DBtable order by email" ;
$user_result = mysql_query($user_sql) or die(mysql_error());
while($user_array = mysql_fetch_array($user_result))
{
$institutea = $user_array[institute] ;
}
$dire = str_replace(" ","-",$institutea);
echo "<br />" . dire; // this shows a value
// and if I do an echo it works but if I do an echo inside the function is does not work, if I do an echo after the function it works
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Or you can write a script that looks up IP address range for specific country's who speck one of the languages and set that language page to load as default and the other for any other IP range
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You can write the name of the second language in its original Codec
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Hello, this is my first post I hope you guys can help me out here, basically whats going on is I declare a variable "$diree" then I insert it in a function but its not getting recognized, here is the code:
$institutea "some directory";
$dire = str_replace(" ","-",$institutea);
# SETTINGS
$max_width = 110;
$max_height = 130;
$per_page = 5;
$page = $_GET['page'];
$has_previous = false;
$has_next = false;
function getPictures() {
global $page, $per_page, $has_previous, $has_next;
if ( $handle = opendir($diree.'/') ) { // done changes here
$lightbox = rand();
?><table border="1"><tr><td>
<form action="pro.php" method="post">
<?php
echo "<ul id='pictures'>";
$count = 1;
$skip = $page * $per_page;
if ( $skip != 0 )
$has_previous = true;
while ( $count < $skip && ($file = readdir($handle)) !== false ) {
if ( !is_dir($file) && ($type = getPictureType($file)) != '' )
$count++;
}
$count = 1;
while ( $count < $per_page && ($file = readdir($handle)) !== false ) {
if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) {
if ( ! is_dir($dire .'/') ) { // done changes here
mkdir($dire .'/'); // done changes here
}
if ( ! file_exists($dire .'/'.$file) ) {
makeThumb( $file, $type );
}
echo '<li>';
echo '<img src="'.$dire .'/'.$file.'" alt="" / width="110" height="110"><br/><input type="checkbox" name="food[]" value='.$file.'>'; // done changes here
echo '</li>';
$count++;
}
}
echo '</ul>';
?>
</td></tr><tr><td><div align="center"><input type="submit" value="Add"></div></td></tr>
</tr></table></form>
<?php
while ( ($file = readdir($handle)) !== false ) {
if ( !is_dir($file) && ($type = getPictureType($file)) != '' ) {
$has_next = true;
break;
}
}
}
}
function getPictureType($file) {
$split = explode($dire .'/', $file);
$ext = $split[count($split) - 1];
if ( preg_match('/jpg|jpeg/i', $ext) ) {
return 'jpg';
} else if ( preg_match('/png/i', $ext) ) {
return 'png';
} else if ( preg_match('/gif/i', $ext) ) {
return 'gif';
} else {
return '';
}
}
function makeThumb( $file, $type ) {
global $max_width, $max_height;
if ( $type == 'jpg' ) {
$src = imagecreatefromjpeg($file);
} else if ( $type == 'png' ) {
$src = imagecreatefrompng($file);
} else if ( $type == 'gif' ) {
$src = imagecreatefromgif($file);
}
if ( ($oldW = imagesx($src)) < ($oldH = imagesy($src)) ) {
$newW = $oldW * ($max_width / $oldH);
$newH = $max_height;
} else {
$newW = $max_width;
$newH = $oldH * ($max_height / $oldW);
}
$new = imagecreatetruecolor($newW, $newH);
imagecopyresampled($new, $src, 0, 0, 0, 0, $newW, $newH, $oldW, $oldH);
if ( $type == 'jpg' ) {
imagejpeg($new, $dire .'/'.$file);
} else if ( $type == 'png' ) {
imagepng($new, $dire .'/'.$file);
} else if ( $type == 'gif' ) {
imagegif($new, $dire .'/'.$file);
}
imagedestroy($new);
imagedestroy($src);
}
?>
small Project I need done in PHP mysql
in Archived Job Offerings
Posted
we need to create a tree type table:
Year --> Institute --> Simister
an admin Page to enter Values
we enter a Year, then we enter multiple Institute affiliated with that Year, then multiple Simisters affiliated with that Institute
the script then will mkdir() where dir name = year-Institute-simister
so if we have 1 year input, 2 Institute inputs and 2 simisters per Institute
we should have 4 folders created.
then I want a display page for users where they Select there Year, then a select menue with avalible Institute's for that year would appear, and when one is selected a Simister list would appear.
can you give me a quote, and time-line for doing it.
almasri_m@hotmail.com